Fork me on GitHub
Math for the people, by the people.

User login

Jacobian determinant

Primary tabs

Jacobian determinant

Hello,

I am trying to derive the formula for converting the differential area dxdy into dudv, where x and y are given as functions of u and v. I saw a derivation (or motivation) in a calculus book which which uses transformation and images, but I wasn't fully convinced as the motivation was not rigorous.

I tried the following, but does not seem to work out:

dx = (\partial x)/(\partial u) du + (\partial x)/(\partial v) dv.

dy = (\partial y)/(\partial u) du + (\partial y)/(\partial v) dv.

When I multiplied them, I did not get the jacobian. I am getting expressions involving du^2 and dv^2, which I do not know what to do with.

Any ideas? Please tell me what's wrong, and would be great if you send a link if you find a rigorous derivation/proof.

Thanks.


I already did! But this is not a derivation. As I mentioned, the only derivation I saw was using transformation and mapping. My question is why doesn't the straightforward way of finding dx and dy in terms of u and v, and multiplying them yield the jacobian??

This is why I got confused.

http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStu...

The above site does have a description of how the determinant of the Jacobian shows up in the change of variables.

It is a little long, but I think it will clarify it. The essential point is that it shows how this determinant is used to change dA from one coordinate system to another. Click on "The Jacobian" to get to the general derivation.

Thanks for the proof. But one should still get the answer when using the equations:

dx = (\partial x)/(\partial u) du + (\partial x)/(\partial v) dv.

dy = (\partial y)/(\partial u) du + (\partial y)/(\partial v) dv.

My problem is how to play around and get rid of du^2 and dv^2.

The image of the rectangle (0,0)-(0,dv)-(du,dv)-(du,0) in the (tangent space of) the u-v plane is a parallelogram in the (tangent space of) the x,y plane with two opposite vertices being(0,0) and (dx,dy). However the area of such a parallelogram is not necessarily dxdy.

In other words the calculus in your argument was fine, but you started off with the wrong expression for the area of a parallelogram.

I understand that an infinitesimal rectangular area dxdy in the x-y plane may not be a rectangle in the u-v plane. But I am simply trying to find a formula relating dxdy to dudv (of course they are not equal). It is just that I am using differentials (instead of transformations and images) to establish this formula, which should end up to be the Jacobian determinant.

There are two ways of looking at it:

Firstly there is the intuitive way that I was referring to before: Let dx, dy, du, dv be the relevant components of tangent vectors. Then your expressions for dx and dy are correct but the key point is that dxdy is not Jdudv:

You have a rectangle in the (tangent space of the) u-v plane with area dudv.

This maps to a parallelogram in the (tangent space of the) x-y plane. You can easily check that the area of this parallelogram is Jdudv and that it has an opposite pair of vertices (0,0) and (dx,dy). Clearly the area Jdudv is in general less than dxdy. I really recommend you draw out this parallelogram and label the vertices to see what is going on.

Now the other way: Let dx, dy, du, dv be the relevant 1-forms. Then dx^dy = J du^dv, (the ^'s are usually not written down).

Your expressions for dx and dy are still correct (basic exercise in dual vector spaces). However when you multiply out dx^dy, you get terms du^du=0, dv^dv=0 by definition of ^.

http://tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx

Try the above. Also google "Jacobean determinant" for more references.

You are not supposed to multiply them together . You are looking for an area and assuming you are in flat space you want to take the determinant or \partial x)/(\partial u) du times (\partial y)/(\partial v) dv - \partial x)/(\partial v) dv times \partial y)/(\partial u) du because in a sense du.du=0 as does dv.dv . The reason for the sign change as in the 2nd term in the determinant is a little obscure and has something to do with right handed or oriented coordinate systems but in a sense may be thought of as a CROSS product which
according to orientation say of two straight lines crossing one into the other it is the x part of line 1 times the y part of line 2 MINUS the y part of line 1 times the x part of line 2 or we could reverse the sign if we crossed line 2 into line 1. Often it is such that don't have to worry about sign and just assume that area is always positive so can just use the absolute value so then the expression fronting du.dv is non-negative.

Subscribe to Comments for "Jacobian determinant"